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zalisa [80]
3 years ago
5

An energy bar contains 9000 milligrams of protein. how much protein does it contain in grams

Mathematics
1 answer:
____ [38]3 years ago
3 0

Answer:

9 Grams

Step-by-step explanation:

You would move the decimal over 3 times to the left  from the end of 9000

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When circuit boards used in the manufacture of compact disc players are tested, the long-run percentage of defectives is 5%. Let
sergiy2304 [10]

Answer:

(a) P(X=3) = 0.093

(b) P(X≤3) = 0.966

(c) P(X≥4) = 0.034

(d) P(1≤X≤3) = 0.688

(e) The probability that none of the 25 boards is defective is 0.277.

(f) The expected value and standard deviation of X is 1.25 and 1.089 respectively.

Step-by-step explanation:

We are given that when circuit boards used in the manufacture of compact disc players are tested, the long-run percentage of defectives is 5%.

Let X = <em>the number of defective boards in a random sample of size, n = 25</em>

So, X ∼ Bin(25,0.05)

The probability distribution for the binomial distribution is given by;

P(X=r)= \binom{n}{r} \times p^{r}\times (1-p)^{n-r}  ; x = 0,1,2,......

where, n = number of trials (samples) taken = 25

            r = number of success

            p = probability of success which in our question is percentage

                   of defectivs, i.e. 5%

(a) P(X = 3) =  \binom{25}{3} \times 0.05^{3}\times (1-0.05)^{25-3}

                   =  2300 \times 0.05^{3}\times 0.95^{22}

                   =  <u>0.093</u>

(b) P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)

= \binom{25}{0} \times 0.05^{0}\times (1-0.05)^{25-0}+\binom{25}{1} \times 0.05^{1}\times (1-0.05)^{25-1}+\binom{25}{2} \times 0.05^{2}\times (1-0.05)^{25-2}+\binom{25}{3} \times 0.05^{3}\times (1-0.05)^{25-3}

=  1 \times 1 \times 0.95^{25}+25 \times 0.05^{1}\times 0.95^{24}+300 \times 0.05^{2}\times 0.95^{23}+2300 \times 0.05^{3}\times 0.95^{22}

=  <u>0.966</u>

(c) P(X \geq 4) = 1 - P(X < 4) = 1 - P(X \leq 3)

                    =  1 - 0.966

                    =  <u>0.034</u>

<u></u>

(d) P(1 ≤ X ≤ 3) =  P(X = 1) + P(X = 2) + P(X = 3)

=  \binom{25}{1} \times 0.05^{1}\times (1-0.05)^{25-1}+\binom{25}{2} \times 0.05^{2}\times (1-0.05)^{25-2}+\binom{25}{3} \times 0.05^{3}\times (1-0.05)^{25-3}

=  25 \times 0.05^{1}\times 0.95^{24}+300 \times 0.05^{2}\times 0.95^{23}+2300 \times 0.05^{3}\times 0.95^{22}

=  <u>0.688</u>

(e) The probability that none of the 25 boards is defective is given by = P(X = 0)

     P(X = 0) =  \binom{25}{0} \times 0.05^{0}\times (1-0.05)^{25-0}

                   =  1 \times 1\times 0.95^{25}

                   =  <u>0.277</u>

(f) The expected value of X is given by;

       E(X)  =  n \times p

                =  25 \times 0.05  = 1.25

The standard deviation of X is given by;

        S.D.(X)  =  \sqrt{n \times p \times (1-p)}

                     =  \sqrt{25 \times 0.05 \times (1-0.05)}

                     =  <u>1.089</u>

8 0
2 years ago
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